Suppose that an arrow is drawn on each edge of a cube, giving each edge a direction, in such a way that every vertex of the cube has at least one arrow coming out of it and at least one arrow going into it. Prove there exists a face of the cube such that the directions of the four boundary edges of that face go in a cycle.
So a cube has 6 faces, 8 vertices, and 12 edges, and there will be 12 arrows in total. each vertex has three edges with one in and two out or one out and two in. WLOG, we can assume one vertex for one in and two out and then we go to the adjacent vertices. Is there an easier way to do this than just exhaust all possibilities by going through all vertices?
The six faces of the cube have in total $24$ right angles. Such an angle is good if it can be traversed according to the drawn arrows, and bad otherwise. It is easy to check that a face can have $0$, $2$, or $4$ bad angles. Now at each vertex of the cube we have two good angles and one bad angle, so that there are $8$ bad angles in total. It follows that there can be at most $4$ faces with $\geq2$ bad angles. This allows to conclude that here are $\geq2$ faces with a directed cycle going around them.